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Question:

If $\cot ^{-1}(\alpha)=\cot ^{-1} 2+\cot ^{-1} 8+\cot ^{-1} 18+\cot ^{-1} 32+\ldots \ldots$ upto 100 terms, then $\alpha$ is :

  1. (1) $1.01$

  2. (2) 1

  3. (3) $1.02$

  4. (4) $1.03$


Correct Option: 1

Solution:

$\operatorname{Cot}^{-1}(\alpha)=\cot ^{-1}(2)+\cot ^{-1}(8)+\cot ^{-1}(18)+\ldots \ldots$

$=\sum_{n=1}^{100} \tan ^{-1}\left(\frac{2}{4 n^{2}}\right)$

$=\sum_{n=1}^{100} \tan ^{-1}\left(\frac{(2 n+1)-(2 n-1)}{1+(2 n+1)(2 n-1)}\right)$

$=\sum_{n=1}^{100} \tan ^{-1}(2 n+1)-\tan ^{-1}(2 n-1)$

$=\tan ^{-1} 201-\tan ^{-1} 1$

$=\tan ^{-1}\left(\frac{200}{202}\right)$

$\therefore \cot ^{-1}(\alpha)=\cot ^{-1}\left(\frac{202}{200}\right)$

$\alpha=1.01$

 

 

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